JEE Advanced Maths – Permutations & Combinations PYQ with Detailed Solutions PDF

Q 1 :

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is [2014]

264
265
53
67

1

2

3

4

(3)

$∵$ Card numbered 1 is always placed in envelope numbered 2, we can consider two cases:

$Case I:$ Card numbered 2 is placed in envelope numbered 1. Then it is derangement of 4 objects, which can be done in

$4! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 9 ways$

$Case II:$ Card numbered 2 is not placed in envelope numbered 1. Then it is derangement of 5 objects, which can be done in

$5! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}) = 44 ways$

$∴$ $Total ways = 44 + 9 = 53$

Q 2 :

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [2009]

55
66
77
88

1

2

3

4

(3)

We have to form 7 digit numbers, using the digits 1, 2 and 3 only, such that the sum of the digits in a number = 10. This can be done by taking 2, 2, 2, 1, 1, 1, 1 or by taking 2, 3, 1, 1, 1, 1, 1.

$∴$ $Number of ways = \frac{7!}{3! 4!} + \frac{7!}{5!} = 77$

Q 3 :

The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is [2007]

360
192
96
48

1

2

3

4

(3)

$The letters of the word C O C H I N in alphabetical order are$ $C, C, H, I, N, O .$

$Fixing first and second letters as C, C, the remaining 4 can be arranged in 4! ways.$

$Similarly, the words starting with each of C H, C I, C N are 4!$

$Then fixing first two letters as C O and next four places when filled in alphabetical order$ $with remaining 4 letters give the word C O C H I N .$

$∴$ $Number of words coming before C O C H I N$

$= 4 \times 4! = 4 \times 24 = 96$

Q 4 :

If the LCM of $p, q$ is $r^{2} t^{4} s^{2}$ , where $r, s, t$ are prime numbers and $p, q$ are positive integers, then the number of ordered pair $(p, q)$ is [2006]

252
254
225
224

1

2

3

4

(3)

$∵$ $r, s, t are prime numbers,$

$∴$ $Section of (p, q) can be done as follows$

[IMAGE 23]

$∴$ $r can be selected in (1 + 1 + 3 = 5) ways$

$Similarly, t and s can be selected in 9 and 5 ways respectively.$

$∴$ $Total number of ordered pairs (p, q) = 5 \times 9 \times 5 = 225$

Q 5 :

The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is [2002]

40
60
80
100

1

2

3

4

(1)

Total number of ways of arranging the letters of the word BANANA is $\frac{6!}{2! 3!} = 60$

Number of words in which 2 N's come together is $\frac{5!}{3!} = 20$

$∴$ $Required number = 60 - 20 = 40$

Q 6 :

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? [2000]

16
36
60
180

1

2

3

4

(3)

$X - X - X - X - X .$ $The four digits 3, 3, 5, 5 can be arranged at (-) places in$

$\frac{4!}{2! 2!} = 6 ways$

$The five digits 2, 2, 8, 8, 8 can be arranged at (X) places in$

$\frac{5!}{3! 2!} = 10 ways$

$∴$ $Total number of arrangements = 6 \times 10 = 60 ways$

Q 7 :

The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is ______. [2018]

(625)

The last 2 digits in a 5-digit number divisible by 4 can be 12, 24, 32, 44 or 52.

Also, each of the first three digits can be any of {1, 2, 3, 4, 5}

$∴$ 5 options for each of the first three digits and total 5 options for last 2 digits

$∴$ Required number of 5-digit numbers are = $5 \times 5 \times 5 \times 5 = 625$

Q 8 :

Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is ______. [2014]

(7)

$∵$ $n_{1}, n_{2}, n_{3}, n_{4} and n_{5} are positive integers such that n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$

$Then for n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$

$If n_{1}, n_{2}, n_{3}, n_{4} take minimum values 1, 2, 3, 4 respectively, then n_{5} will be maximum 10 .$

$∴$ $Corresponding to n_{5} = 10, there is only one solution$

$n_{1} = 1, n_{2} = 2, n_{3} = 3, n_{4} = 4 .$

$Corresponding to n_{5} = 9, we can have only one solution$

$n_{1} = 1, n_{2} = 2, n_{3} = 3, n_{4} = 5 i.e., one solution.$

$Corresponding to n_{5} = 8, we can have, only solution$

$n_{1} = 1, n_{2} = 2, n_{3} = 3, n_{4} = 6$

$or n_{1} = 1, n_{2} = 2, n_{3} = 4, n_{4} = 5$

$i.e., 2 solutions.$

$For n_{5} = 7, we can have$

$n_{1} = 1, n_{2} = 2, n_{3} = 4, n_{4} = 6$

$or n_{1} = 1, n_{2} = 3, n_{3} = 4, n_{4} = 5$

$i.e., 2 solutions.$

$For n_{5} = 6, we can have$

$n_{1} = 2, n_{2} = 3, n_{3} = 4, n_{4} = 5$

$i.e., one solution.$

$∴$ $Thus there can be 7 solutions.$

Q 9 :

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ______. [2020]

(1080)

Groups can be possible in only 2, 2, 1, 1 way.

Number of ways of dividing persons in group

$= \frac{6!}{{(2!)}^{2} {(1!)}^{2} {(2!)}^{2}}$

Number of ways after arranging rooms $= \frac{6!}{{(2!)}^{4}} \cdot 4! = 1080$

Q 10 :

Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ____. [2019]

(30)

5 persons A, B, C, D and E are seated in circular arrangement.
Let A be given red hat, then there will be two cases.

Case I: B and E have same coloured hat blue/green. Say B and E have blue hat.
Then C and D can have either red and green or green and red i.e. 2 ways.

[IMAGE 24]---------------

Similarly if B & E have green hat, there will be 2 ways for C & D.
Hence there are 2 + 2 = 4 ways.

Case II: B and E have different coloured hats blue and green or green and blue.

[IMAGE 25]-------------------

Let B has blue and E has green.
If C has green then D can have red or blue.
If C has red then D can have only blue.
∴ three ways.

Similarly 3 ways will be there when B has green and E has blue.
∴ there are 3 + 3 = 6 ways.

On combining the two cases, there will be 4 + 6 = 10 ways.
When similar discussion is repeated with A as blue and green hat, we get 10 ways for each.
Therefore, in all, there will be 10 + 10 + 10 = 30 ways.

Q 11 :

The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ______. [2022]

(569)

[IMAGE 26]

Q 12 :

Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS. [2008]

Column I Column II

(A) The number of permutations containing the word ENDEA is (p) 5!

(B) The number of permutations in which the letter E occurs in the first and the last positions is (q) 2 × 5!

(C) The number of permutations in which none of the letters D, L, N occurs in the last five positions is (r) 7 × 5!

(D) The number of permutations in which the letters A, E, O occur only in odd positions is (s) 21 × 5!

$(A) \to (q); (B) \to (s); (C) \to (q); (D) \to (p)$
$(A) \to (q); (B) \to (q); (C) \to (s); (D) \to (p)$
$(A) \to (s); (B) \to (q); (C) \to (q); (D) \to (p)$
$(A) \to (p); (B) \to (s); (C) \to (q); (D) \to (q)$

1

2

3

4

(4)

(A) For the permutations containing the word ENDEA we consider 'ENDEA' as single letter. Then we have total ENDEA, N, O, E, L i.e. 5 letters which can be arranged in 5! ways.

$∴$ $(A) \to (p)$

(B) If E occupies the first and last position, the middle 7 positions can be filled by N, D, E, A, N, O, L in
$\frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 21 \times 120 = 21 \times 5!$ ways.

$∴$ $(B) \to (s)$

(C) If none of the letters D, L, N occur in the last five positions then we should arrange D, L, N, N at first four positions and rest five i.e. E, E, E, A, O at last five positions. This can be done in
$\frac{4!}{2!} \times \frac{5!}{3!} = 4 \times 3 \times \frac{5!}{3 \times 2} = 2 \times 5!$ ways.

$∴$ $(C) \to (q)$

(D) As per question A, E, E, E, O can be arranged at 1st, 3rd, 5th, 7th and 9th positions and rest D, L, N, N at remaining 4 positions. This can be done in
$\frac{5!}{3!} \times \frac{4!}{2!} ways = 2 \times 5!$ ways.

$∴$ $(D) \to (q)$

Topic Question Set

Q 2 :

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [2009]

Q 3 :

The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is [2007]

Q 4 :

If the LCM of $p, q$ is $r^{2} t^{4} s^{2}$ , where $r, s, t$ are prime numbers and $p, q$ are positive integers, then the number of ordered pair $(p, q)$ is [2006]

Q 5 :

The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is [2002]

Q 6 :

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? [2000]

Q 7 :

The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is ______. [2018]

Q 8 :

Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is ______. [2014]

Q 9 :

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ______. [2020]

Q 10 :

Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ____. [2019]

Q 11 :

The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ______. [2022]

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	Column I		Column II
(A)	The number of permutations containing the word ENDEA is	(p)	5!
(B)	The number of permutations in which the letter E occurs in the first and the last positions is	(q)	2 × 5!
(C)	The number of permutations in which none of the letters D, L, N occurs in the last five positions is	(r)	7 × 5!
(D)	The number of permutations in which the letters A, E, O occur only in odd positions is	(s)	21 × 5!

Topic Question Set

Q 2 : The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [2009]

Q 3 : The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is [2007]

Q 4 : If the LCM of p,q is r2t4s2, where r,s,t are prime numbers and p,q are positive integers, then the number of ordered pair (p,q) is [2006]

Q 5 : The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is [2002]

Q 6 : How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? [2000]

Q 7 : The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is ______. [2018]

Q 8 : Let n1<n2<n3<n4<n5 be positive integers such that n1+n2+n3+n4+n5=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5) is ______. [2014]

Q 9 : In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ______. [2020]

Q 10 : Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ____. [2019]

Q 11 : The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ______. [2022]

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Q 2 :

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [2009]

Q 3 :

The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is [2007]

Q 4 :

If the LCM of $p, q$ is $r^{2} t^{4} s^{2}$ , where $r, s, t$ are prime numbers and $p, q$ are positive integers, then the number of ordered pair $(p, q)$ is [2006]

Q 5 :

The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is [2002]

Q 6 :

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? [2000]

Q 7 :

The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is ______. [2018]

Q 8 :

Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is ______. [2014]

Q 9 :

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ______. [2020]

Q 10 :

Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ____. [2019]

Q 11 :

The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ______. [2022]